Question

In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.​

Answer 1

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$\Large\fbox{\color{purple}{QUESTION}}$

In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

➡️In any triangle ABC,

➡️a/sin A = b/sin B = c/sin C = k

➡️a = k sin A, b = k sin B, c = k sin C

✴➡️LHS

➡️= a sin (B – C) + b sin (C – A) + c sin (A – B)

➡️= k sin A [sin B cos C – cos B sin C] + k sin B [sin C cos A – cos C sin A] + k sin C [sin A cos B

– cos A sin B]

➡️= k sin A sin B cos C – k sin A cos B sin C + k sin B sin C cos A – k sin B cos C sin A + k sin C sin A cos B – k sin C cos A sin B

➡️= 0

➡️= RHS

➡️Hence proved that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.

$\Large\mathcal\green{FOLLOW \: ME}$

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Answer 2

Answer:

⏩ SOLUTION:-

➡️In any triangle ABC,

➡️a/sin A = b/sin B = c/sin C = k

➡️a = k sin A, b = k sin B, c = k sin C

✴➡️LHS

➡️= a sin (B – C) + b sin (C – A) + c sin (A – B)

➡️= k sin A [sin B cos C – cos B sin C] + k sin B [sin C cos A – cos C sin A] + k sin C [sin A cos B

– cos A sin B]

➡️= k sin A sin B cos C – k sin A cos B sin C + k sin B sin C cos A – k sin B cos C sin A + k sin C sin A cos B – k sin C cos A sin B

➡️= 0

➡️= RHS

➡️Hence proved that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.✔

Step-by-step explanation:

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